//给定一个二叉树，判断它是否是高度平衡的二叉树。
//
// 本题中，一棵高度平衡二叉树定义为：
//
//
// 一个二叉树每个节点 的左右两个子树的高度差的绝对值不超过 1 。
//
//
//
//
// 示例 1：
//
//
//输入：root = [3,9,20,null,null,15,7]
//输出：true
//
//
// 示例 2：
//
//
//输入：root = [1,2,2,3,3,null,null,4,4]
//输出：false
//
//
// 示例 3：
//
//
//输入：root = []
//输出：true
//
//
//
//
// 提示：
//
//
// 树中的节点数在范围 [0, 5000] 内
// -104 <= Node.val <= 104
//
// Related Topics 树 深度优先搜索 二叉树
// 👍 1457 👎 0


//leetcode submit region begin(Prohibit modification and deletion)


import java.util.Map;

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public boolean isBalanced(TreeNode root) {

        return getHeight(root) == -1 ? false : true;
    }
    /*
    我的想法
    public int getHeight(TreeNode node) {
        if (left == null && right == null) {
            return true;
        }
        int leftDepth = 0;
        int rightDepth = 0;
        while (left.left != null) {
            leftDepth++;
        }
        while (right.right != null) {
            rightDepth++;
        }
        if (Math.abs(leftDepth - rightDepth) > 1) {
            return false;
        } else {
            return true;
        }
        boolean judgeLeft = judge_Balance(left.left, left.right);
        boolean judgeRight = judge_Balance(right.left, right.right);
        return judgeLeft && judgeRight;
    }*/

    public int getHeight(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int leftHeight = getHeight(root.left);  //左
        if (leftHeight == -1)  return -1;
        int rightHeight = getHeight(root.right);  //右
        if (rightHeight == -1)  return -1;
        int result;
        if (Math.abs(leftHeight - rightHeight) > 1) {
            return -1;
        } else {
            result = 1 + Math.max(leftHeight, rightHeight);
        }
        return result;
    }
}
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